The height of a ball thrown vertically upward from a rooftop is modelled by h(t)=-5t^2=20t=50, where h(t) is the ball's height above the gorund,in meters, at time t seconds after the throw
a) determine the maximum height of the ball
b) how long does it take for the ball to reach its maximum height
c) how highis the rooftop
Read the problem carefully and if you still don't understand let me know :D
Your equation makes no sense, make sure you read what you typed before posting.
To find the maximum height of the ball, first note that the equation H(t) = -5t^2 + 20t + 50 represents the height of the ball above the ground at time t seconds.
a) To determine the maximum height, we need to find the vertex of the parabolic equation H(t). The vertex of a parabola in the form ax^2 + bx + c is given by the formula t = -b/2a. In this case, a = -5 and b = 20. Plugging these values into the formula, we get t = -20/(2*(-5)) = -20/(-10) = 2.
Now substitute t = 2 back into the original equation to find the value of H(2):
H(2) = -5(2)^2 + 20(2) + 50 = -20 + 40 + 50 = 70.
Therefore, the maximum height of the ball is 70 meters.
b) To find how long it takes for the ball to reach its maximum height, we already know that it reached the maximum height at t = 2 seconds.
c) To determine the height of the rooftop, we need to find the value of H(0).
H(0) = -5(0)^2 + 20(0) + 50 = 50.
Therefore, the height of the rooftop is 50 meters.